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On the other hand, if the $G$-structure is a complex structure on $M$, all of these $G$-structures are locally equivalent, so they are locally homogeneous to all orders. Now, when $M$=\mathbb{CP}^2$, the biholomorphisms act transitively on the complex frame bundle, but when $M=\mathbb{CP}^1\times \mathbb{CP}^1$, the biholomorphisms act transitively on $M$ but not on its complex frame bundle.

On the other hand, if the $G$-structure is a complex structure on $M$, all of these $G$-structures are locally equivalent, so they are locally homogeneous to all orders. Now, when $M=\mathbb{CP}^2$, the biholomorphisms act transitively on the complex frame bundle, but when $M=\mathbb{CP}^1\times \mathbb{CP}^1$, the biholomorphisms act transitively on $M$ but not on its complex frame bundle.

Your question, then, easily reduces to "Is the homomorphism $D:\mathrm{Diff}(M,\Omega,x)\to \mathrm{SL}(T_xM)$ defined by $D(f) = f'(x):T_xM\to T_xM$ surjective?"

The answer is 'yes'. The reason is that the image of $D$ has to be a Lie subgroup of $\mathrm{SL}(T_xM)$, and it is very easy to see that it contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$. Consequently, it must be all of $\mathrm{SL}(T_xM)$.

Added at the request of the OP: Showing that the image of the homomorphism $D$ contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$ can be done in a number of ways, but here is one that is relatively 'low-tech': Note that the result is obvious when $n=1$, so assume that $n>1$ from now on. Then, one can choose coordinates $u=(u^i)$ centered on $x\in M$ and defined on a neighborhood $U$ of $x$ in which $$ \Omega = \mathrm{d}u^1\wedge\mathrm{d}u^2\wedge\cdots\wedge\mathrm{d}u^n = \mathrm{d}u $$ There is an $r>0$, so that $u(U)\subset\mathbb{R}^n$ contains the open ball $B_r(0)$ of radius $r$ about $0\in\mathbb{R}^n$. Let $G_0 = \mathrm{Diff}_0\bigl(B_r(0),\mathrm{d}u,0\bigr)$ denote the group of diffeomorphisms $\phi:B_r(0)\to B_r(0)$ that satisfy $\phi(0)=0$, $\phi^*(\mathrm{d}u) = \mathrm{d}u$, and for which there exists a compact set $K_\phi\subset B_r(0)$ such that $\phi$ is the identity outside the compact set $K_\phi$. Clearly, it will be enough to show that the homomorphism $D:G_0\to \mathrm{SL}\bigl(T_0B_r(0)\bigr) = \mathrm{SL}(\mathbb{R}^n)$ is surjective.

Your question, then, reduces to "Is the homomorphism $D:\mathrm{Diff}(M,\Omega,x)\to \mathrm{SL}(T_xM)$ defined by $D(f) = f'(x):T_xM\to T_xM$ surjective?"

The answer is 'yes'. The reason is that the image of $D$ has to be a Lie subgroup of $\mathrm{SL}(T_xM)$, and a simple local construction (see the remark at the end) shows that it must be all of $\mathrm{SL}(T_xM)$.

As for your more general question, this has been considered at length in the literature, beginning with the work of Élie Cartan on what are now called 'pseudo-groups' and continuing with a very extensive development in the 1950s and 1960s by Chern, Kuranishi, Singer, Sternberg, Guillemin, and many others. The basic question is this: "If an $n$-manifold is endowed with a $G$-structure $B\subset \mathcal{F}(M)$ (where $G\subset\mathrm{GL}(n,\mathbb{R})$ is a subgroup and $\mathcal{F}(M)$ is the (co-)frame bundle of $M$, when does $\mathrm{Diff}(M,B)$, the group of diffeomorphisms of $M$ that preserve $B$, act transitively on $M$ and when does it act transitively on $B$?" This latter transitivity is a very strict condition, and it can hold 'locally' without holding globally.

For example, when the $G$-structure is a Riemannian metric, $\mathrm{Diff}(M,B)$ acts transitively on $M$ iff $M$ is homogeneous as a Riemannian manifold, but $\mathrm{Diff}(M,B)$ acts transitively on $B$ iff $M$ has constant sectional curvature and is globally symmetric.

On the other hand, if the $G$-structure is a complex structure on $M$, all of these $G$-structures are locally equivalent, so they are locally homogeneous to all orders. Now, when $M$=\mathbb{CP}^2$, the biholomorphisms act transitively on the complex frame bundle, but when $M=\mathbb{CP}^1\times \mathbb{CP}^1$, the biholomorphisms act transitively on $M$ but not on its complex frame bundle.

As a final example, when the $G$-structure is a symplectic structure on $M$ (and $M$ is connected), the symplectomorphisms do act transitively on both $M$ and $B$.

Thus, the answer to your question depends very much on which $G$-structure you want to study.

Remark added at the request of the OP: Showing that the image of the homomorphism $D$ contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$ can be done in a number of ways, but here is one that is relatively 'low-tech': Note that the result is obvious when $n=1$, so assume that $n>1$ from now on. Then, one can choose coordinates $u=(u^i)$ centered on $x\in M$ and defined on a neighborhood $U$ of $x$ in which $$ \Omega = \mathrm{d}u^1\wedge\mathrm{d}u^2\wedge\cdots\wedge\mathrm{d}u^n = \mathrm{d}u $$ There is an $r>0$, so that $u(U)\subset\mathbb{R}^n$ contains the open ball $B_r(0)$ of radius $r$ about $0\in\mathbb{R}^n$. Let $G_0 = \mathrm{Diff}_0\bigl(B_r(0),\mathrm{d}u,0\bigr)$ denote the group of diffeomorphisms $\phi:B_r(0)\to B_r(0)$ that satisfy $\phi(0)=0$, $\phi^*(\mathrm{d}u) = \mathrm{d}u$, and for which there exists a compact set $K_\phi\subset B_r(0)$ such that $\phi$ is the identity outside the compact set $K_\phi$. Clearly, it will be enough to show that the homomorphism $D:G_0\to \mathrm{SL}\bigl(T_0B_r(0)\bigr) = \mathrm{SL}(\mathbb{R}^n)$ is surjective.

Let $M$ be an $n$-manifold endowed with a nonvanishing $n$-form $\Omega$, let $\mathrm{Diff}(M,\Omega)$ denote the group of $\Omega$-preserving diffeomorphisms of $M$, and, for $x\in M$, let $\mathrm{Diff}(M,\Omega,x)$ denote the subgroup that fixes $x$.

Your question, then, easily reduces to "Is the homomorphism $D:\mathrm{Diff}(M,\Omega,x)\to \mathrm{SL}(T_xM)$ defined by $D(f) = f'(x):T_xM\to T_xM$ surjective?"

The answer is 'yes'. The reason is that the image of $D$ has to be a Lie subgroup of $\mathrm{SL}(T_xM)$, and it is very easy to see that it contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$. Consequently, it must be all of $\mathrm{SL}(T_xM)$.

Added at the request of the OP: Showing that the image of the homomorphism $D$ contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$ can be done in a number of ways, but here is one that is relatively 'low-tech': Note that the result is obvious when $n=1$, so assume that $n>1$ from now on. Then, one can choose coordinates $u=(u^i)$ centered on $x\in M$ and defined on a neighborhood $U$ of $x$ in which $$ \Omega = \mathrm{d}u^1\wedge\mathrm{d}u^2\wedge\cdots\wedge\mathrm{d}u^n = \mathrm{d}u $$ There is an $r>0$, so that $u(U)\subset\mathbb{R}^n$ contains the open ball $B_r(0)$ of radius $r$ about $0\in\mathbb{R}^n$. Let $G_0 = \mathrm{Diff}_0\bigl(B_r(0),\mathrm{d}u,0\bigr)$ denote the group of diffeomorphisms $\phi:B_r(0)\to B_r(0)$ that satisfy $\phi(0)=0$, $\phi^*(\mathrm{d}u) = \mathrm{d}u$, and for which there exists a compact set $K_\phi\subset B_r(0)$ such that $\phi$ is the identity outside the compact set $K_\phi$. Clearly, it will be enough to show that the homomorphism $D:G_0\to \mathrm{SL}\bigl(T_0B_r(0)\bigr) = \mathrm{SL}(\mathbb{R}^n)$ is surjective.

To do this, consider the subset $C\subset \mathrm{Hom}_0(\mathbb{R}^n,\mathbb{R}^n)={\frak{sl}}(n,\mathbb{R})$ consisting of the operators with trace zero whose eigenvalues are purely imaginary and either all nonzero (if $n$ is even) or one zero with multiplicity 1 (if $n$ is odd). Note that $C$ is an open cone in ${\frak{sl}}(n,\mathbb{R})$. It is the interior of the set of operators $L:\mathbb{R}^n\to\mathbb{R}^n$ for which the $1$-parameter subgroup $\mathrm{e}^{tL}$ has compact closure in $\mathrm{SL}(n,\mathbb{R})$. In particular, for such an operator, there is at least one positive definite inner product $\langle,\rangle$ on $\mathbb{R}^n$, such that the flow of $\mathrm{e}^{tL}$ preserves the inner product $\langle,\rangle$ and hence the volume form $\mathrm{d}u$. Now, let $\epsilon>0$ be so small that the set of vectors $y\in\mathbb{R}^n$ that satisfy $\langle y,y\rangle \le \epsilon$ is a compact subset of $B_r(0)$. Let $h:\mathbb{R}\to\mathbb{R}$ be a smooth function that is identically $1$ when $0\le t \le \epsilon/2$ and vanishes identically for $t\ge\epsilon$. Now consider the $1$-parameter family of smooth maps $$ f_t(y) = \mathrm{e}^{t\ h(\langle y,y\rangle) L}y. $$ This family preserves the level sets of $Q(y) = \langle y,y\rangle$, which are spheres, and rigidly rotates each one, so it preserves the volume form $\mathrm{d}u$. It clearly lies in $G_0$. When $Q(y)\ge\epsilon$, $f_t(y) = y$, and, when $Q(y)\le \epsilon/2$, we have $f_t(y) = \mathrm{e}^{tL}y$. Thus, $f_t$ lies in $G_0$. Thus, since $$ D(f_t) = \mathrm{e}^{tL}, $$ it follows that the image of $D$ contains the subset $\mathrm{e}^{C}\subset \mathrm{SL}(n,\mathbb{R})$, which obviously has non-empty interior. This subset is also closed under inverse, since $C = -C$, and from this, it follows that $\mathrm{e}^{C}$ generates $\mathrm{SL}(n,\mathbb{R})$. In particular, it follows that the image of $D$ is all of $\mathrm{SL}(n,\mathbb{R})$, which is what needed to be proved.

Let $M$ be an $n$-manifold endowed with a nonvanishing $n$-form $\Omega$, let $\mathrm{Diff}(M,\Omega)$ denote the group of $\Omega$-preserving diffeomorphisms of $M$, and, for $x\in M$, let $\mathrm{Diff}(M,\Omega,x)$ denote the subgroup that fixes $x$.

Your question, then, easily reduces to "Is the homomorphism $D:\mathrm{Diff}(M,\Omega,x)\to \mathrm{SL}(T_xM)$ defined by $D(f) = f'(x):T_xM\to T_xM$ surjective?"

The answer is 'yes'. The reason is that the image of $D$ has to be a Lie subgroup of $\mathrm{SL}(T_xM)$, and it is very easy to see that it contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$. Consequently, it must be all of $\mathrm{SL}(T_xM)$.

Added at the request of the OP: Showing that the image of the homomorphism $D$ contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$ can be done in a number of ways, but here is one that is relatively 'low-tech': Note that the result is obvious when $n=1$, so assume that $n>1$ from now on. Then, one can choose coordinates $u=(u^i)$ centered on $x\in M$ and defined on a neighborhood $U$ of $x$ in which $$ \Omega = \mathrm{d}u^1\wedge\mathrm{d}u^2\wedge\cdots\wedge\mathrm{d}u^n = \mathrm{d}u $$ There is an $r>0$, so that $u(U)\subset\mathbb{R}^n$ contains the open ball $B_r(0)$ of radius $r$ about $0\in\mathbb{R}^n$. Let $G_0 = \mathrm{Diff}_0\bigl(B_r(0),\mathrm{d}u,0\bigr)$ denote the group of diffeomorphisms $\phi:B_r(0)\to B_r(0)$ that satisfy $\phi(0)=0$, $\phi^*(\mathrm{d}u) = \mathrm{d}u$, and for which there exists a compact set $K_\phi\subset B_r(0)$ such that $\phi$ is the identity outside the compact set $K_\phi$. Clearly, it will be enough to show that the homomorphism $D:G_0\to \mathrm{SL}\bigl(T_0B_r(0)\bigr) = \mathrm{SL}(\mathbb{R}^n)$ is surjective.

To do this, consider the subset $C\subset \mathrm{Hom}_0(\mathbb{R}^n,\mathbb{R}^n)={\frak{sl}}(n,\mathbb{R})$ consisting of the operators that are skew-symmetric with respect to some positive definite inner product on $\mathbb{R}^n$. Let $L\in C$ be such an operator, and let $\langle,\rangle$ on $\mathbb{R}^n$ be a positive definite inner product with respect to which $L$ is skew-symmetric. In particular, the $1$-parameter subgroup $\mathrm{e}^{tL}$ preserves $\langle,\rangle$ and hence the volume form $\mathrm{d}u$. Now, let $\epsilon>0$ be so small that the set of vectors $y\in\mathbb{R}^n$ that satisfy $\langle y,y\rangle \le \epsilon$ is a compact subset of $B_r(0)$. Let $h:\mathbb{R}\to\mathbb{R}$ be a smooth function that is identically $1$ when $0\le t \le \epsilon/2$ and vanishes identically for $t\ge\epsilon$. Now consider the $1$-parameter family of smooth maps $$ f_t(y) = \mathrm{e}^{t\ h(\langle y,y\rangle) L}\ y. $$ This family preserves the level sets of $Q(y) = \langle y,y\rangle$, which are spheres, and isometrically rotates each one, so it preserves the volume form $\mathrm{d}u$. It clearly lies in $G_0$. When $Q(y)\ge\epsilon$, $f_t(y) = y$, and, when $Q(y)\le \epsilon/2$, one has $f_t(y) = \mathrm{e}^{tL}y$. Note that $f_t$ lies in $G_0$. Since $$ D(f_t) = \mathrm{e}^{tL}, $$ it follows that the image of $D$ contains the subset $\mathrm{e}^{C}\subset \mathrm{SL}(n,\mathbb{R})$. In particular, the image of $D$, which is a subgroup of $\mathrm{SL}(n,\mathbb{R})$, contains all of the compact subgroups of $\mathrm{SL}(n,\mathbb{R})$. Thus, it follows that the image of $D$ is all of $\mathrm{SL}(n,\mathbb{R})$, which is what needed to be proved.